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I have an observation in my real analysis lecture notes that states that if $f,g:\,(X,\mu)\to\mathbb{R}$ (with Borel's -$\sigma$ algebra ) and $f=g$ almost everywhere then if $g$ is Lebesgue measurable then so if $f$.

I don't understand why this is true, can someone please explain ?

I tried looking at some Borel set and on it source, but I can't figure why if we change $g$ in some measure $0$ of points then the source is still in Lebesgue -$\sigma$ algebra

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1 Answer 1

up vote 4 down vote accepted

Let $A$ be a measurable subset of $\mathbb R$. We must show that $Y=f^{-1}(A)$ is measurable in $X$.

By hypothesis, the set

$$ N=\lbrace x \in X | f(x) \neq g(x) \rbrace $$ is measurable and has measure $0$. We deduce that $Y \cap N$ is also measurable with measure $0$. So it suffices to show that $Y\setminus N$ is measurable.

Now $Y\setminus N =g^{-1}(A) \setminus N$ is a difference of two measurable sets, qed.

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Thanks, got it! –  Belgi Dec 30 '12 at 7:47

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